1
$\begingroup$

(1) Calculation of Max. and Min. value of $\sqrt{x^2-3x+2}+\sqrt{2+3x-x^2}\;\; \forall x\in \mathbb{R}$

(2) Calculation of Max. value of $\sqrt{10x-x^2}-\sqrt{18x-x^2-77}\;\;\forall x\in \mathbb{R}$

My Try:: for $(1)$ one:: Using the Cauchy-Schwarz inequality

$$\Rightarrow \displaystyle \left\{\left(\sqrt{x^2-3x+2}\right)^2+\left(\sqrt{2+3x-x^2}\right)^2\right\}\cdot \left(1^2+1^2\right)\geq \left\{\sqrt{x^2-3x+2}+\sqrt{2+3x-x^2}\right\}^2$$ $$\Rightarrow \displaystyle \left\{\sqrt{x^2-3x+2}+\sqrt{2+3x-x^2}\right\}^2\leq 8$$

$$\Rightarrow \left\{\sqrt{x^2-3x+2}+\sqrt{2+3x-x^2}\right\}\leq 2\sqrt{2}$$

and equality hold when $\displaystyle \sqrt{x^2-3x+2}=\sqrt{2+3x-x^2}\Rightarrow x=0\;,3$

Is my solution right for Max.? If not then please help me, and how can I calculate for Min. and also help required in $(2)$ one.

$\endgroup$
  • $\begingroup$ It's correct, you can check it here: wolframalpha.com/input/… $\endgroup$ – Zafer Cesur Jan 7 '14 at 14:49
  • $\begingroup$ Too bad you cannot use derivatives. Your answer is correct. $\endgroup$ – user114628 Jan 7 '14 at 14:49
  • $\begingroup$ Why can't we use derivatives BUT are allowed to assume knowledge of Cauchy-Schwarz inequality!? $\endgroup$ – Squirtle Jan 7 '14 at 15:06
  • $\begingroup$ Since you classified this as "algebra-precalculus" (despite the mention of Cauchy-Schwartz inequality), I thought I'd try precalculus methods. Completing the square under the radicals in (1) leads to $\sqrt{u^2 + \frac{1}{4}} + \sqrt{-u^2 + \frac{17}{4}},$ where $u = x - \frac{3}{2}$ and $-\frac{1}{2}\sqrt{17} \leq u \leq \frac{1}{2}\sqrt{17}$ (the restriction on $u$ is from $-u^2 + \frac{17}{4} \geq 0$). At this point I don't see how to continue using only precalculus methods. (continued) $\endgroup$ – Dave L. Renfro Jan 7 '14 at 15:06
  • $\begingroup$ (continuation) However, it's now an easy first semester "max/min on a closed and bounded interval" problem, with the max/min candidates winding up being $u = 0, \; \pm \sqrt{2}, \; \pm \frac{1}{2}\sqrt{17}.$ $\endgroup$ – Dave L. Renfro Jan 7 '14 at 15:07
1
$\begingroup$

(1):the approach is not right ,you can't make sure $a\ge b,b\le c \to a \ge c $

the easy way is let$ u=x^2-3x, f(x)=f(u)=\sqrt{2+u}+\sqrt{2-u},f^2=4+2\sqrt{4-u^2}$

$0\le\sqrt{4-u^2} \le 2 \implies 4 \le f^2 \le 8 \implies 2 \le f \le 2\sqrt{2}$ when $u=0$ and $u=2$ get Max and Min.

(2): $\sqrt{10x-x^2}$ domain is [$0,10$], $\sqrt{18x-x^2-77}$ domain is [$7,11$] so $f(x)=\sqrt{10x-x^2}-\sqrt{18x-x^2-77}$ domain is [$7,10$]

$\sqrt{10x-x^2}$ is mono decreasing on [$7,10$], $\sqrt{18x-x^2-77}$ is increasing on [$7,9$] so we can sure $f(x)$ will get Max when $x=7$ when $x$ on [$7,9$], the remain is [$9,10$],but it is impossible for any $f(x) \ge f(7)$ because it is always $ <f(7)- a, a >0$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.